A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women.

Solution:

Here, Number of men = 10

Number of women = 7

6 committee numbers can be selected containing at-least 3 men and 2 women in following 2 ways.

→ 4 men and 2 women 

→ 3 men and 3 women

∴ number of ways of selecting at-least 3 men and 2 women in committee of 6

$={ }^{10} C_{3} \times{ }^{7} C_{3}+{ }^{10} C_{4}+{ }^{7} C_{2}$

$=\frac{10 \times 9 \times 7}{3 \times 2} \times \frac{7 \times 6 \times 5}{6}+\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2} \times \frac{7 \times 6}{2}$

$=120 \times 35+210 \times 21$

$=8610$

Number of ways when two particular women are never together = 10C3

$=\frac{{ }^{10} C_{4}}{\text { all men }}+\frac{{ }^{10} C_{3} \times{ }^{5} C_{1}}{2 \text { women }}$

$=210+120 \times 5$

$=210+600$

$=810$

Therefore, Total number of ways when two particular women are never together = 8610 − 810 = 7800.